zoukankan      html  css  js  c++  java
  • 电梯时间计算


    // Z_1008.cpp : 定义控制台应用程序的入口点。
    //2015-07-13
    /*
    The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.
    
    For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.
     即电梯调度算法
     */
    
    #include<iostream>
    using namespace std;
    
    int* sort(int *a,int N)
    {
    	int top = 0,j=1;
    	int *b = new int[N + 1];
    	b[0] = 0;
    	for (int i = 0; i < N; i++)
    	{
    		if (top < a[i])
    		{
    			top = a[i];
    			a[i] = 0;
    			b[j] = top;
    			j++;
    		}
    	}
    	for (int i = 1; i < N; i++)
    		for (int m = 0; m < N-i;m++)
    		{
    			if (a[m] < a[m + 1])
    			{
    				int n = a[m]; a[m] = a[m + 1]; a[m + 1] = n;
    			}
    		}
    		top = j;
    	for (int i = 0; i <= N-top ; i++)
    	{
    		b[j] = a[i];
    		j++;
    	}
    	return b;
    }
    
    int time(int *b,int N)
    {
    	int sum = 0;
    	for (int i = 0; i < N; i++)
    	{
    		int x = b[i] - b[i + 1];
    		if (x<0)
    		{
    			sum += (-6 * x + 5);
    		}
    		else
    		{
    			sum += (4 * x + 5);
    		}
    	}
    	return sum;
    }
    
    int main(void)
    {
    	int N=111,j=0,ans[256];
    	while (N > 0)
    	{
    		cout << "input require number:";
    		cin >> N;
    		int  *a=new int[N], buff, *b;
    		for (int i = 0; i < N; i++)
    		{
    			cin >> buff;
    			if (buff>0 && buff <= 100)
    				a[i] = buff;
    			else
    				cout << "erro:overside!
    ";
    		}
    		b = sort(a, N);
    		ans[j]=time(b, N);
    		j++;
    	}
    	for (int i = 0; i < j-1; i++)
    	{
    		cout << ans[i] << endl;
    	}
    
    }
    
    /*
     *  1.动态数组的申请
     *	2.指针的活用
    */

  • 相关阅读:
    装饰器 如何理解Python装饰器?
    python装饰器详解
    window 10下安装jdk
    Linux中pam认证详解(上)
    VMware/KVM/OpenStack虚拟化之网络模式总结
    java问题排查命令
    今日面试问题
    Qwtplot3D Qt5.12.0 MinGW 编译运行
    20212022年寒假学习进度06
    Springboot笔记<10>常用注解总结
  • 原文地址:https://www.cnblogs.com/A-yes/p/9894263.html
Copyright © 2011-2022 走看看